Subnetting a Class B Address with a 6-bit Subnet Field

Given: the address of a Class B host is to be divided into subnets using a 6-bit subnet number.

A classful Class B address has 16 network bits and 16 host bits:

| network id : 16 bits | host id : 16 bits |
| network id : 16 bits | subnet : 6 bits | host : 16 - 6 = 10 bits |

1. Maximum number of subnets

Subnet bits are borrowed from the host id, so

$$N_{subnets} = 2^{6} = \mathbf{64}\ \text{subnets}$$

(If the all-zeros subnet and the all-ones subnet are not usable, as in the old classful rules, only $64 - 2 = 62$ subnets are usable, but the maximum count is $2^6 = 64$.)

2. Maximum number of hosts in each subnet

Host bits left after borrowing 6 bits:

$$h = 16 - 6 = 10\ \text{bits}$$

$$N_{hosts} = 2^{10} - 2 = 1024 - 2 = \mathbf{1022}\ \text{hosts per subnet}$$

The subtraction of 2 removes the subnet's network address (all host bits 0) and its broadcast address (all host bits 1).

3. Number of networks allowed under Class B addresses

Class B reserves a 2-bit class prefix (10), leaving a 14-bit network id:

$$N_{networks} = 2^{14} = \mathbf{16384}\ \text{Class B networks}$$

Final answers: maximum subnets = 64, maximum hosts per subnet = 1022, number of Class B networks = 16384.